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tiny-mole [99]
2 years ago
15

During a physics experiment, a class drops a golf

Mathematics
2 answers:
sergey [27]2 years ago
4 0
I did the math and the answer of your question is 2.17 seconds.
Anit [1.1K]2 years ago
3 0

Answer:

2.17 seconds

brainliest pls

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The average lifetime of a certain new cell phone is 3.4 years. The manufacturer will replace any cell phone failing within three
melisa1 [442]

Answer:

68% of these phones last 3.87 years.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The average lifetime of a certain new cell phone is 3.4 years.

This means that m = 3.4, \mu = \frac{1}{3.4} = 0.2941

So

P(X \leq x) = 1 - e^{-0.2941x}

68% of these phones last how long (in years)?

This is x for which:

P(X \leq x) = 0.68

P(X \leq x) = 1 - e^{-0.2941x}

Then

0.68 = 1 - e^{-0.2941x}

e{-0.2941x} = 0.32

\ln{e{-0.2941x}} = \ln{0.32}

-0.2941x = \ln{0.32}

x = -\frac{\ln{0.32}}{0.2941}

x = 3.87

68% of these phones last 3.87 years.

4 0
3 years ago
How do i find the ratio between 16 and 20
Nookie1986 [14]

Answer:

divide 16/20

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Help pleaseeeeeeee!!!!!!!!
alina1380 [7]

Answer:

4,3

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Brainlist for the right answer no link or bot 30 points
Ber [7]

Answer:

what is the question

Step-by-step explanation:

there is no question

7 0
3 years ago
Helpp!!!!<br> Evaluate (x+5)^2+8 for x=-2<br> A. 12<br> B. 33<br> C. 17<br> D. 57
VashaNatasha [74]
Answer: C. 17

Explanation:
(x+5)^2+8
(-2x+5)^2+8
3^2 + 8
9 + 8
= 17
6 0
3 years ago
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